Question 1160833
the point on the line {{{y=4x}}} that is closest to the point {{{P}}}=({{{1}}},{{{2}}}) will be intersection point of the line {{{y=4x}}} and the line perpendicular to {{{y=4x}}} that passes through the point {{{P}}}=({{{1}}}},{{{2}}}) 

the line {{{y=4x}}} have a slope {{{m=4}}}

and the perpendicular line will have a slope negative reciprocal which is {{{-1/4}}}

then, the equation of the perpendicular line will be

{{{y=-(1/4)x+b}}}......use given point {{{P}}}=({{{1}}}},{{{2}}}) to calculate {{{b}}}

{{{2=-(1/4)1+b}}}

{{{2+(1/4)=b}}}

{{{9/4=b}}}

the equation of the perpendicular line is:

{{{y=-(1/4)x+9/4}}}


now find intersection point {{{I}}} solving the system:

{{{y=4x}}}.........eq.1

{{{y=-(1/4)x+9/4}}}....eq.2
-----------------------------------------
{{{4x=-(1/4)x+9/4}}}

{{{4x+(1/4)x=9/4}}}....both sides multiply by {{{4}}}

{{{16x+x=9}}}

{{{17x=9}}}

{{{x=9/17}}}-> {{{x}}} coordinate

{{{y=4(9/17)=36/17}}}->{{{y}}} coordinate

intersection point: {{{I}}}=({{{9/17}}},{{{36/17}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,2,.12),circle(9/17,36/17,.12),
locate(1,2.5,P),locate(9/17,2,I),
 graph( 600, 600, -10, 10, -10, 10, 4x,-(1/4)x+9/4)) }}}